Question

The threshold frequency V⁰ for a metal is 7.0 x 10¹⁴s. Calculate the kinetic energy of an electron emitted when radiation of frequency V=1.0x 10¹⁵s⁻ hits the metal

Answer 1

If frequency of radiation = 1.0 x 10¹⁵s⁻¹
Energy of radiation is equal to Planck's Constant(6.626 x 10⁻³⁴ m² kg / s) multiplied by frequency
Energy of radiation=6.626 x 10⁻¹⁹ J

An electron on the surface requires a certain amount of energy to free itself from the atom. That energy is (Threshold frequency x Planck's constant)
Energy to free electron=4.638 x 10⁻¹⁹ J

Kinetic Energy of electron= Energy of radiation - Energy to free electron

KE=1.988 x 10⁻¹⁹ J ≈ 2 x 10⁻¹⁹ J

Answer 2

K.E ( kinetic energy ) = h ( f - f₀ ) 

           h ----- > plank's constant

           f₀ ------> Threshold frequency

           f --------> Frequency 

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         h = 6.626 × 10⁻³⁴

         f₀ = 7× 10¹⁴ Hz

         f = 1 × 10¹⁵ Hz

K.E = (6.626 ×10⁻³⁴ ) ₓ [ (1 × 10¹⁵)  - (7× 10¹⁴)]

      = 1.987 × 10⁻¹⁹ J //